I read in a book that the number of solutions of x+y+z=11 where x,y,z belong to [1,6] and are integers, can be given by coefficient of \(x^{11}\) in expansion of \((x+x^2+x^3+.....+x^6)^3\). I can't understand the method. Can someone explain the logic behind it? And is there a general statement for this?

## Comments

Sort by:

TopNewestConsider using the distributive property to write the following product as the sum of a bunch of terms:

\((t^1+t^2+t^3+t^4+t^5+t^6)(t^1+t^2+t^3+t^4+t^5+t^6)(t^1+t^2+t^3+t^4+t^5+t^6)\).

If you pick the \(t^x\) term from the first sum, the \(t^y\) term from the first sum, and the \(t^z\) term from the first sum, then the product will be \(t^{x+y+z}\).

So, each solution \((x,y,z)\) to \(x+y+z = 11\) with \(x,y,z \in \overline{1,6}\) yields a \(t^{11}\) term.

Thus, the \(t^{11}\) coefficient of the expansion is the number of such solutions.

This method goes under the category of Generating Functions. There are several sources online that formalize this and give more examples. – Jimmy Kariznov · 4 years ago

Log in to reply

– Shubham Srivastava · 4 years ago

Oh! It was so simple. Thanks for explaining.Log in to reply

I am also not able to understand this theorem.I am also stuck with it.I think here multinomial theorem is used. – Kishan K · 4 years ago

Log in to reply